A certain water tank has two outlet valves, valve A and valve B. Each

Question

A certain water tank has two outlet valves, valve A and valve B. Each valve drains water from the tank at a constant rate. If the tank is full and valve B alone is opened, does it take more than two hours to completely drain the tank?

(1) When the tank is full, if valve A alone is opened, it takes 4 hours to completely drain the tank.

(2) When the tank is full, if valve A is opened and valve B is opened an hour later, it takes 4/3 as long to completely drain the tank as it would if both valves had been opened simultaneously from the start.

Bunuel · Accepted Answer

The question has a typo. It should be  it takes   4/3   as long to completely drain the tank

A certain water tank has two outlet valves, valve A and valve B. Each valve drains water from the tank at a constant rate. If the tank is full and valve B alone is opened, does it take more than two hours to completely drain the tank?

(1) When the tank is full, if valve A alone is opened, it takes 4 hours to completely drain the tank. No info on valve B. Not sufficient.

(2) When the tank is full, if valve A is opened and valve B is opened an hour later, it takes 4/3 as long to completely drain the tank as it would if both valves had been opened simultaneously from the start.

Say  the rate  of valve A is  a  tank/hour and  the rate  of valve B is  b  tank/hour.

In 1 hour valve A would drain (rate)(time) = a*1 = a part of the tank. So, 1 - a part of the tank will be left to drain. To drain 1 - a part of the tank, both A and B will need  (time) = \frac{(job)}{(rate)} = \frac{(1 - a)}{(a + b)}  hours. So, the total time in this case is  1 +\frac{(1 - a)}{(a + b)}  hours.

Next, to drain the full tank from the start, both A and B will need  (time) = \frac{(job)}{(rate)} = \frac{1}{(a + b)}  hours.

We are told that:  1 + \frac{(1 - a)}{(a + b)} = \frac{4}{3}*\frac{1}{(a + b)} .

\frac{3(b+1)}{a+b}=\frac{4}{a+b} ;

From this we can get that  b = \frac{1}{3}  tank/hour. So, the time it would take valve B to drain the tank is 3 hours. Sufficient.

Answer: B.

kunal20 · Answer

Hi, Can you pls tel how total time was derived ? 1+(1−a)/(a+b) , I got the logic behind (1-a)/(a+b), but not 1 + (1-a)/(a+b)

Bunuel · Answer

"Valve A is opened and valve B is opened  an hour later , ..." So, for an hour valve A drained the tank alone. After B joined together they needed (1−a)/(a+b) hours, hence total of 1 + (1−a)/(a+b) hours.

kunal20 · Answer

Got it. Thanks Bunuel

amanvermagmat · Answer

Thankyou Bunuel for noticing the mistake and editing this one. I was pretty confused. Heres how I did this:

Let valve A drains 'a' units per hour, let valve B drains 'b' units per hour, and let the water in tank be total 'a*b' units. 
So valve A will take = ab/a = b hours, while valve B will take = ab/b = a hours : to drain water on their own each. We need to know whether a > 2 or not?

(1) valve A alone takes 4 hours or b=4. or total water in tank = 4a units. But this doesnt give the answer to our question. Insufficient.

(2) if both valves are opened simultaneously, units of water drained per hour = (a+b), thus time taken = ab/(a+b)
IF instead first A is opened for 1 hour, 'a' units of water already drained in one hour. Units left in tank = (ab-a) ...now both valves are opened and units being drained per hour = a+b, thus further time now required to drain completely = (ab-a)/(a+b). So total time required in this case = 1 + (ab-a)/(a+b) OR (b+ab)/(a+b)

We are given that this latter time is 4/3 or the former time, so we can write this as:
(b+ab)/(a+b) = 4/3 * ab/(a+b) ... Solving we get: a+1 = 4a/3 Or a=3. So we get that a>2 (or valve B will take more than 2 hours to drain the tank). Sufficient.

Hence  B answer

TheNightKing · Answer

From Statement B, you can see that the only different is B opening one hour later and causing delay (or lets say adding) 1/3 more time.
Which means to add delay of 100% it will take 3 hours. In other words it will take 3 hours to empty the tank.

GMATGuruNY · Answer

Statement 1:
No info about B.
INSUFFICIENT.

Statement 2:
Let the tank = 3 gallons, A's hourly rate = 1, and B's hourly rate = B, implying that the combined rate for A and B together = 1+B

Since the combined rate for A and B = 1+B, the time for A and B together to empty the 3-gallon tank  = \frac{work}{combined-rate} = \frac{3}{1+B}

If A works on its own for an hour -- emptying 1 gallon of the 3-gallon tank -- the time for A and B together to empty the remaining 2 gallons  = \frac{remaining-work}{combined-rate} = \frac{2}{1+B} , with the result that the total number of work-hours to empty the tank = A's time alone + time for A and B together =  1 + \frac{2}{1+B}

Since the second time is equal to 4/3 of the first time, we get:
 1 + \frac{2}{1+B }= \frac{4}{3}*\frac{3}{1+B} 
 \frac{1+B+2}{1+B} = \frac{4}{1+B} 
 B+3 = 4 
 B = 1 
Since B empties 1 gallon per hour, the time for B on its own to empty the 3-gallon tank =  \frac{work}{rate} = \frac{3}{1} = 3  hours
Thus, the answer to the question stem is YES.
SUFFICIENT.

B

Suruchim12 · Answer

GMATGuruNY

I have a question with regards to the approach yo have shared. In that, we have assumptions made at 2 steps ; 1 where we assume the tank capacity to be 3 gallons and next when we assume that Valve A when it works on its own for an hour -- empties 1 gallon of the 3-gallon tank.

While the first assumption is understandable. the 2nd assumption isnt something that I'd think of in the exams if I was faced with this problem. Therefore, Is there an alternative approach to solving this question?

GMATGuruNY · Answer

One option is to test a second value for A's rate to confirm that we get the same result as when A=1.

Let the tank = 3 gallons, A's hourly rate = 2, and B's hourly rate = B, implying that the combined rate for A and B together = 2+B

Since the combined rate for A and B = 2+B, the time for A and B together to empty the 3-gallon tank  = \frac{work}{combined-rate} = \frac{3}{2+B}

If A works on its own for an hour -- emptying 2 gallons of the 3-gallon tank-- the time for A and B together to empty the remaining 1 gallon  = \frac{remaining-work}{combined-rate} = \frac{1}{2+B} , with the result that the total number of work-hours to empty the tank = A's time alone + time for A and B together =  1 + \frac{1}{2+B}

Since the second time is equal to 4/3 of the first time, we get:
 1 + \frac{1}{2+B }= \frac{4}{3}*\frac{3}{2+B} 
 \frac{2+B+1}{2+B} = \frac{4}{A+B} 
 B+3 = 4 
 B = 1 
Since B empties 1 gallon per hour, the time for B on its own to empty the 3-gallon tank =  \frac{work}{rate} = \frac{3}{1} = 3  hours

B takes 3 hours whether A=1 or A=2.
Thus, the answer to the question stem is YES.
SUFFICIENT.

Plugging in a value for A's rate allows us to perform the algebra using only one unknown (B).
Alternatively, we can leave A's rate as a variable.

Let the tank = 3 gallons, A's hourly rate = A, and B's hourly rate = B, implying that the combined rate for A and B together = A+B

Since the combined rate for A and B = A+B, the time for A and B together to empty the 3-gallon tank  = \frac{work}{combined-rate} = \frac{3}{A+B}

If A works on its own for an hour -- emptying A gallons of the 3-gallon tank -- the time for A and B together to empty the remaining 3-A gallons  = \frac{remaining-work}{combined-rate} = \frac{3-A}{A+B} , with the result that the total number of work-hours to empty the tank = A's time alone + time for A and B together =  1 + \frac{3-A}{A+B}

Since the second time is equal to 4/3 of the first time, we get:
 1 + \frac{3-A}{A+B }= \frac{4}{3}*\frac{3}{A+B} 
 \frac{A+B+3-A}{A+B} = \frac{4}{A+B} 
 B+3 = 4 
 B = 1 
Since B empties 1 gallon per hour, the time for B on its own to empty the 3-gallon tank =  \frac{work}{rate} = \frac{3}{1} = 3  hours
Thus, the answer to the question stem is YES.
SUFFICIENT.

agrasan · Answer

Let's say  L  be the volume of tank, and  A litres/hour  and  B litres/hour  be respective rates for valves A and B

(1) When the tank is full, if valve A alone is opened, it takes 4 hours to completely drain the tank.

L/A = 4, this is clearly  insufficient  as it doesn't give us any info the rate of B

(2) When the tank is full, if valve A is opened and valve B is opened an hour later, it takes 4/3 as long to completely drain the tank as it would if both valves had been opened simultaneously from the start.

If we translate this statement, A is remained open for 1 hour and then A+B remained opened for sometime
In 1 hour, volume of water drained out by A alone = A*1 = A litres
Volume of water remaining = L - A
Time taken by A+B together to completely drain out remaining water = (L-A)/(A+B)

Now, we are given total time is 4/3 times of time taken when A+B started working  right from the beginning

1 + (L-A)/(A+B) = (4/3)*(L/(A+B))
Simplifying, we get B = L/3, this means it takes 3 hours for B alone to drain out the tank completely

Thus, statement(2)  alone is sufficient

A certain water tank has two outlet valves, valve A and valve B. Each

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A certain water tank has two outlet valves, valve A and valve B. Each

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